Question 17.4: A hollow square duct of the configuration shown (left) has i...
A hollow square duct of the configuration shown (left) has its surfaces maintained at 200 and 100 \mathrm{~K}, respectively. Determine the steady-state heat transfer rate between the hot and cold surfaces of this duct. The wall material has a thermal conductivity of 1.21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}. We may take advantage of the eightfold symmetry of this figure to lay out the simple square grid shown below (right).
The grid chosen is square with \Delta x=\Delta y=1 / 2 \mathrm{~m}. Three interior node points are thus identified; their temperatures may be determined by proper application of equation (17-63).
T_{i,j}={\frac{T_{i-1,j}+T_{i+1,j}+T_{i,j-1}+T_{i,j+1}}{4}} (17.63)
Writing the proper expressions for T_{1}, T_{2}, and T_{3} using equation (17-68) as a guide, we have
\begin{aligned} & T_{1}=\frac{200+100+2 T_{2}}{4} \\ & T_{2}=\frac{200+100+T_{1}+T_{3}}{4} \\ & T_{3}=\frac{100+100+2 T_{2}}{4} \end{aligned}
This set of three equations and three unknowns may be solved quite easily to yield the following: T_{1}=145.83 \mathrm{~K}, T_{2}=141.67 \mathrm{~K}, T_{3}=120.83 \mathrm{~K}.
The temperatures just obtained may now be used to find heat transfer. Implicit in the procedure of laying out a grid of the sort, we have specified is the assumption that heat flows in the x and y directions between nodes. On this basis heat transfer occurs from the hot surface to the interior only to nodes 1 and 2; heat transfer occurs to the cooler surface from nodes 1,2, and 3. We should also recall that the section of duct that has been analyzed is one-eighth of the total thus, of the heat transfer to and from node 1, only one half should be properly considered as part of the element analyzed.
We now solve for the heat transfer rate from the hotter surface, and write
\begin{aligned} q & =\frac{k\left(200-T_{1}\right)}{2}+k\left(200-T_{2}\right) \\ & =k\left[\left(\frac{200-145.83}{2}\right)+(200-141.67)\right] \\ & =85.415 k \quad(q \text { in } \mathrm{W} / \mathrm{m}, k \text { in } \mathrm{W} / \mathrm{m} \cdot \mathrm{K}) \end{aligned}
A similar accounting for the heat flow from nodes 1,2 , and 3 to the cooler surface is written as
\begin{aligned} q & =\frac{k\left(T_{1}-100\right)}{2}+k\left(T_{2}-100\right)+k\left(T_{3}-100\right) \\ & =k\left[\left(\frac{145.83-100}{2}\right)+(141.67-100)+(120.83-100)\right] \\ & =85.415 k \quad(q \text { in } \mathrm{W} / \mathrm{m}, k \text { in } \mathrm{W} / \mathrm{m} \cdot \mathrm{K}) \end{aligned}
Observe that these two different means of solving for q yield identical results. This is obviously a requirement of the analysis and serves as a check on the formulation and numerical work.
The example may now be concluded. The total heat transfer per meter of duct is calculated as
\begin{aligned} q & =8(8.415 \mathrm{~K})(1.21 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}) \\ & =826.8 \mathrm{~W} / \mathrm{m} \end{aligned}
